Şekildeki gibi, şeklimizi kareye tamamlayalım.
$[DF] \perp [BE]$ ve $[DG] \perp [CE]$
$|DF|=|DG|=|GE|=|FE|=x$
$|BF|=x \sqrt{3}$
$|AE|=|BE|\sqrt{2}=x(1+\sqrt{3})\sqrt{2}$
$|AD|=x(1+\sqrt{3})\sqrt{2}-x\sqrt{2}=x\sqrt{6}=6 \rightarrow x=\sqrt{6}$
$A(\widehat{BDC})=A(\widehat{BCE})-2.A(\widehat{BDE})=\frac{|BE|.|CE|}{2}-|DF|.|BE|=\frac{(x(1+\sqrt{3}))^2}{2}-(x^2(1+\sqrt{3}))=6$
